# Divergence of Electric Field Due to a Point Charge [duplicate]

I am trying to formally learn electrodynamics on my own (I only took an introductory course). I have come across the differential form of Gauss's Law.

$$\nabla \cdot \mathbf E = \frac {\rho}{\epsilon_0}.$$

That's fine and all, but I run into what I believe to be a conceptual misunderstanding when evaluating this for a point charge.

I know the math looks better in spherical coordinates, but I will be using Cartesian.

So when I calculate the divergence I obtain:

$$\nabla \cdot \mathbf E = \nabla \cdot kQ\langle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\rangle = \frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

This can further be simplified:

$$\frac{-3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}+\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{3-3}{(x^2+y^2+z^2)^{\frac{3}{2}}}.$$

Now instinctively I would say that 3-3 is zero and then the while thing is zero everywhere. I am confused as to why (purely mathematically) this expression is not equal to zero at the origin. I completely understand why it physically has to be that way. And I also understand that it is modeled with the delta dirac function. But what (again, mathematically) is stopping me from saying that equation is just zero even at the origin?

• Hint: You are applying the rules of differentiation where the field is not defined/singular/not differentiable. Jan 9, 2018 at 10:21
• Yes! Very true. Thank you. I should have thought of that.
– Ben
Jan 9, 2018 at 11:24
• It is not customary to use angle brackets to denote components of a vector. Better use () or []. Jan 10, 2019 at 14:14
• I picked up the angle bracket habit from mathematicians.
– Ben
Feb 19, 2019 at 23:11

What you want to compute is essentially $$\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3}$$ at the origin. Of course, that doesn't exist as a function since the field is singular. On the other hand, you have already shown that it vanishes everywhere else.
So you need to interpret the expression in a weak sense, i.e. as a distribution, and consider the integral $$\int_{B_\epsilon}\vec\nabla \,\frac{\vec x}{\left|\vec x\right|^3} \,\text{d}^3x$$ over some volume containing the origin, conveniently chosen as a ball of radius $\epsilon$, convert it to a surface integral which does not include the singularity and see that the result is finite.
• @Ben: No, I'm using it in the simple sense of "function defined at every point in space": The electric field assigns a vector to every point $x$, i.e. it is a function of a variable $x\in \mathbb{R}^3$, and in this case, the function is sigular and not differentiable at $x=0$. Jan 11, 2018 at 12:04